Derivative of cot²x

We now understand the derivative of cot²x. It is commonly represented as d/dx (cot²x) or (cot²x)', and its value is -2cot(x)csc²(x). The function cot²x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Cotangent Function: (cot(x) = cos(x)/sin(x)). Quotient Rule: Rule for differentiating cot(x) (since it consists of cos(x)/sin(x)). Cosecant Function: csc(x) = 1/sin(x).

The derivative of cot²x can be denoted as d/dx (cot²x) or (cot²x)'. The formula we use to differentiate cot²x is: d/dx (cot²x) = -2cot(x)csc²(x) (or) (cot²x)' = -2cot(x)csc²(x) The formula applies to all x where sin(x) ≠ 0

We can derive the derivative of cot²x using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of cot²x results in -2cot(x)csc²(x) using the above-mentioned methods: By First Principle The derivative of cot²x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of cot²x using the first principle, we will consider f(x) = cot²x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = cot²x, we write f(x + h) = cot²(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [cot²(x + h) - cot²x] / h = limₕ→₀ [ [cos²(x + h)/sin²(x + h)] - [cos²x/sin²x] ] / h = limₕ→₀ [ [cos(x + h)sin(x) - sin(x + h)cos(x)] / [sin²(x) · sin²(x + h)] ]/ h We now use the formula cos A sin B - sin A cos B = -sin (A - B). f'(x) = limₕ→₀ [-sin (x + h - x)] / [ h sin²(x) · sin²(x + h)] = limₕ→₀ [-sin h] / [ h sin²(x) · sin²(x + h)] = limₕ→₀ (-sin h)/ h · limₕ→₀ 1 / [sin²(x) · sin²(x + h)] Using limit formulas, limₕ→₀ (-sin h)/ h = -1. f'(x) = -1 [ 1 / (sin²(x) · sin²(x + 0))] = -1/sin⁴x As the reciprocal of sine is cosecant, we have, f'(x) = -csc²(x)cot(x)csc(x). Hence, proved. Using Chain Rule To prove the differentiation of cot²x using the chain rule, We use the formula: cot x = cos x/ sin x Consider f(x) = cos x and g (x)= sin x So we get, cot x = f (x)/ g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1) Let’s substitute f(x) = cos x and g (x) = sin x in equation (1), d/dx (cot x) = [(-sin x)(sin x)- (cos x)(cos x)]/ (sin x)² (-sin²x - cos²x)/ sin²x …(2) Here, we use the formula: (sin²x) + (cos²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (cot x) = -1/ (sin x)² Since csc x = 1/sin x, we write: d/dx(cot x) = -csc²x Using Product Rule We will now prove the derivative of cot²x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, cot²x = (cot x)² Given that, u = cot x and v = cot x Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = d/dx (cot x) = -csc²x (substitute u = cot x) v = cot x (substitute v = cot x) v' = d/dx (cot x) = -csc²x Again, use the product rule formula: d/dx (cot²x) = u'·v + u·v' Let’s substitute u = cot x, u' = -csc²x, v = cot x, and v' = -csc²x When we simplify each term: We get, d/dx (cot²x) = -csc²x·cot x - cot x·csc²x = -2cot(x)csc²x Thus: d/dx (cot²x) = -2cot(x)csc²x.

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like cot²(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′ (x) Similarly, the third derivative, f′′′(x) is the result of the second derivative and this pattern continues. For the nth Derivative of cot²(x), we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).

When the x is 0, the derivative of cot²x = -2cot(0)csc²(0), which is undefined because cot(0) is undefined. When the x is π, the derivative of cot²x = -2cot(π)csc²(π), which is 0 because cot(π) is 0.

Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Cotangent Function: The cotangent function is one of the primary six trigonometric functions and is written as cot x. Cosecant Function: A trigonometric function that is the reciprocal of the sine function. It is typically represented as csc x. Quotient Rule: A method used to differentiate functions that are the ratio of two differentiable functions. Chain Rule: A rule used to differentiate composite functions, expressing the derivative of a composite function as the product of the derivatives of its inner and outer functions.